For values of class numbers of quadratic number fields used below one can see [1]}, [2]}.

If \(p\)  and \(q\)  satisfy conditions (REF ), then by class number formula (cf. [3]}) we have
\(h_2(F)&=&\frac{1}{2^5}q(F)h_2(p)h_2(q)h_2(-p)h_2(-q)h_2(pq)h_2(-pq)h_2(-1), \\&=&\frac{1}{2^5}q(F)\cdot 1\cdot 1\cdot 2\cdot 1 \cdot 2 \cdot 2 \cdot 1=\frac{1}{2^2}q(F).\) 
Since, by Lemma REF , a FSU of \(F^+=\mathbb {Q}(\sqrt{q}, \sqrt{p})\)  is given by \(\lbrace \varepsilon _{p}, \varepsilon _{q}, \sqrt{\varepsilon _{q}\varepsilon _{pq}}\rbrace \) 
and \(\sqrt{2\varepsilon _{q}}\)  is a square in \(F^+\) , then by Lemma REF , \(\lbrace \varepsilon _{p}, \sqrt{\varepsilon _{q}\varepsilon _{pq}}, \sqrt{i\varepsilon _{q}}\rbrace \) 
is a FSU of \(F\) . Thus \(h_2(F)=\frac{1}{2^2}\cdot 4=1\) .

If \(p\)  and \(q\)  satisfy conditions \( {(\ref {cond 2})}\) , then as above we get
\(h_2(F)&=&\frac{1}{2^5}q(F)h_2(p)h_2(q)h_2(-p)h_2(-q)h_2(pq)h_2(-pq)h_2(-1), \\&=&\frac{1}{2^5}q(F)\cdot 1\cdot 1\cdot 1\cdot 1 \cdot 1 \cdot 4 \cdot 1=\frac{1}{2^3}q(F).\) 
By Remark REF  we have \( \sqrt{2\varepsilon _{q}}\) , \(\sqrt{2\varepsilon _{p}}\in F^+\) , so it is easy to see that a FSU of \(F^+\)  is
\(\lbrace \varepsilon _{q}, \sqrt{\varepsilon _{p}\varepsilon _{q}}, \sqrt{\varepsilon _{p}\varepsilon _{pq}} \text{ or } \sqrt{\varepsilon _{pq}}\rbrace \)  or
\(\lbrace \varepsilon _{q}, {\varepsilon _{pq}}, \sqrt{\varepsilon _{p}\varepsilon _{q}}\rbrace \) . Thus, by Lemma REF  and the fact that \(\frac{1}{2^3}q(F)\in \mathbb {N}\) , we have \(q(F)=2^3\)  (and thus \(\lbrace \varepsilon _{q}, {\varepsilon _{pq}}, \sqrt{\varepsilon _{p}\varepsilon _{q}}\rbrace \)  is not a FSU of \(F^+\) ). Therefore \(h(F)\)  is odd.

